Beyond the Golden Ratio

[THEME MUSIC] This is the golden ratio, phi.

It gets a lot of attention.

But what about the silver ratio?

[THEME MUSIC] Cut a line segment into unequal pieces of lengths a and b such as the ratio a to b is the same as the ratio a plus b to a.

That is so that the whole big segment over the medium piece equals the medium piece over the small piece.

This is how you construct the golden ratio, phi.

If a rectangle has an aspect ratio of phi, then you can subdivide it forever into a square and another golden rectangle at each level.

And you can make fun logarithmic spirals by connecting the corners.

Now instead, let's make two medium pieces and one small piece out of our original segment, such that once again big over medium equals medium over small.

This is how you construct the silver ratio or silver mean.

You can even play the same rectangle and spiral game with two squares at each level of subdivision.

OK, now keep incrementing the number of medium pieces or squares in the rectangle picture while always requiring that big over medium equal medium over small.

You then produce a whole sequence of so-called metallic ratios or metallic means, the nth one of which all denote as sigma n. Now the golden ratio hogs all the press, so you may be less familiar with its mathematical metallic cousins.

And that's unfortunate, because many of the sexy mathematical features commonly attributed solely to phi actually have equally cool analogs in the rest of the metallic ratio family.

And that's why we're doing this episode to showcase these analogs so that you can explore them further if you feel inclined.

But please watch all the way through to the end, because I'm going to end with a feature of phi that I'm not quite sure does generalize to the other segments and I'm hoping some of you will be able to help figure it out.

Let's start with the defining equation for the golden ratio.

It can be rewritten as phi equals 1 over phi plus 1.

In other words, the golden ratio is one more than its reciprocal.

With a little algebra we see that this condition is equivalent to the quadratic equation phi squared minus phi minus 1 equals 0, which has two conjugate solutions, one of which is positive, one negative.

Now since phi is the ratio of two lengths, it has to be the positive solution.

That's 1 plus the square root of 5 all over 2.

But the conjugate solution isn't exactly a throw away.

Notice that phi plus its conjugate equals 1.

And going back to our original equation, that phi minus its reciprocal also equals 1.

So apparently the conjugate of phi is the negative reciprocal of phi.

OK. All of the above properties generalize.

Watch.

The defining equation for the silver ratio sigma 2 can also be rewritten as sigma 2 equals 1 over sigma 2 plus 2.

So the silver ratio is two more than its reciprocal.

More generally, sigma n is n more than its reciprocal.

Again, you can encode this condition into a quadratic equation that differs from phi's quadratic equation only in that its linear coefficient is now minus n instead of minus 1.

And once again, you find two conjugates solutions of opposite sides.

Sigma n will be the positive one.

That works out to n plus the square root of n squared plus 4 all over 2.

And again, sigma n's conjugate turns out to its negative reciprocal.

Here's the table of all of these values for the first few n just for reference.

Now let's connect this to some number theory.

One way to represent real numbers is with continued fractions, which are basically nested layers of fractions inside fractions.

For example, pi can be written as 3 plus 1 over 7 plus 1 over 15 plus 1 over 1 plus 1 over 292 plus and so on forever, since the continued fraction of an irrational number is infinite.

Rational numbers, in contrast, have finite continued fractions.

For instance, 1 plus 1 over 2 plus 1 over 3 plus 1 over 2.

Now as long as we agree to use only 1s as numerators, continued fractions are unique.

But some continued fractions look cooler than others.

For example, 1 plus 1 over plus 1 over 1 plus 1 over 1 plus and so on, just infinite 1s all the way down the chain.

What number x does this correspond to?

Well notice that if we zoom in on the first denominator and blow it up, it looks identical to the entire continued fraction.

So whatever x is, it's apparently equal to 1 plus 1 over x.

But wait?

That's just the defining equation for phi.

That's pretty neato.

OK, what if we had 2s all the way down?

This would represent a number that is two more than its reciprocal by similar logic, i.e.

the silver ratio, sigma 2.

Now extrapolate.

The nth metallic ratio sigma n must therefore be the number whose simple continued fraction consists of ns all the way down the chain.

And here we see yet another example of a property that is not specific to phi.

Now there's a cool connection between phi's continued fraction and the Fibonacci sequence, each term of which is the sum of the previous two starting with 1 and 1 as seed terms.

The sequence of ratios of successive Fibonacci terms is well known to tend toward phi in the infinite limit.

But why does that happen?

Well, consider a geometric sequence in which each successive term is exactly phi times the previous term.

So for example, 1 then phi, phi squared, phi cubed, and so forth.

Notice that the third term, phi squared, is exactly the sum of the previous two terms.

Just look back at the quadratic equation that phi satisfies.

If you keep multiplying that equation by phi, you see that every term in this sequence is the sum of the previous two.

In other words, a geometric sequence whose term ratio is phi will automatically satisfy the Fibonacci recurrence relation.

So any sequence satisfying this same recurrence relation will get closer and closer to being geometric with phi as the term ratio.

This includes not just the Fibonacci sequence, but any sequence satisfying this same recurrence relation, regardless of the starting value.

Like for instance, the Lucas numbers.

But among all such sequences, there is something quite special about the Fibonacci sequence specifically.

Namely, go back to the continued fraction for phi and cut it off after one nesting level, and then two nesting levels, then three nesting levels, and so forth.

The terms you get are the best rational approximations to phi with that denominator or lower.

And low and behold, those are precisely the ratios of successive Fibonacci terms.

But once again, this coolness is not the sole purview of phi.

Let's imagine a geometric sequence with a term ratio of sigma n. Looking back at the defining equation for the nth metallic ratio, we can see that each term in this sequence will be n times the previous term plus the term before that.

So now let's make a Fibonacci-like sequence that satisfies this recurrence relation with seed terms of 1 and n instead of 1 and 1.

And let's list the ratios of its successive terms.

Once again, you get the successive finite cut offs of sigma n's continued fraction, which is pretty cool.

So that's my whirlwind tour of the metallic ratios.

We touched on some geometry, some algebra, and some number theory.

Here's a summary of the properties that we discussed.

But of course, that just begins to scratch the surface.

There is a lot of fun structure to explore here, and I encourage you to do so.

I myself started toying around with the following question, which I have not yet managed to answer conclusively.

So the golden ratio is known to feature prominently in the regular pentagon, specifically the ratio of the diagonal of a pentagon to its side is phi.

Now the silver ratio, sigma 2, makes a similar cameo appearance in the regular octagon.

It's the ratio of the octagon's second diagonal to its side.

But what about beyond that?

Are any of the sigma ns equal to any diagonal to side ratio in any regular polygon?

I've coded up a crude numerical exploration for sigma 3 and found nothing so far.

But what I'd really love is either a proof that it never happens or an explicit example of when it does happen for at least one sigma n. Now this is an open ended question.

I don't know the answer.

But if any of you feel inclined to search and come up with one.

Then A, publish it, and B, please email us to let us know.

We'll shout you out in a future episode and send you some PBS swag.

Until then, good metallic hunting and I'll see you guys soon.